Page:Carroll - Euclid and His Modern Rivals.djvu/299

Rh

Now let AB, CD make with AC two interior ∠s BAC, ACD together < 2 right ∠s. It shall be proved that they will meet if produced towards B, D.

In CD take any point D. Join AD. At A make ∠DAH equal to ∠CBA.

Hence AH, CD are equally inclined to all transversals;

∴ ∠s HAC, ACD together = 2 right ∠s;

∴ they together > ∠s BAC, ACD;

∴ ∠HAC > ∠BAC, i.e. ∠HAD > ∠BAD;

∴ ∠CDA > ∠BAD.

At D, in Line DA, make an ∠ equal to ∠BAD;

then the Line, so drawn, will fall within ∠CDA, and will meet CA between C and A. Call this point K.

In CA produced take CL a multiple of CK, and > CA. And in CD produced take CM the same multiple of CD that CL is of CK. And join LM.

Because Triangles CKD, CLM have a common vertical ∠, and the 2 sides of the one respectively equimultiples of those of the other,

∴, by what has been already proved, ∠s CKD, CLM are equal.

Because AB, KD are equally inclined to AD,

∴ they are equally inclined to CL;

∴ ∠CAB = ∠CKD i.e. = ∠CLM;

∴ AB is separational from LM;

∴ if produced, it will meet CM.

Therefore a Pair of Lines, &c.